A 4x4 matrix representation of SU(3)?

You might have a look at Fulton's book Young tableaux for the combinatorics relevant for computing these dimensions. Here is the relevant fact: the irreducible representations of $SU(n)$ may be indexed by partitions with at most $n-1$ parts in such a way that the dimension of the irreducible $L(\lambda)$ corresponding to a partition $\lambda$ is the number of column-strict Young tableaux (in the alphabet $\{1,2,\dots,n\}$) on $\lambda$. This gives a practical way of computing the dimensions that is a bit faster than using Weyl's character formula (the classical formula that can be expressed in a root-system uniform fashion) or more recent tools such as the Littelmann path model. The point being, for $SU(n)$ you don't need to know about root systems to operate a machine that will compute the dimensions. (In fact, things are even better: Schur functions give you the characters of the irreps and not just their dimensions).

In your example, the irreps would therefore be indexed by the partitions (in roughly increasing order of size) $$(0),(1),(1,1),(2),(2,1),(3),(2,2),(3,1),(4),\dots$$ of dimensions $$1,3,3,6,8,10,6,15,15,... $$ as noted in Jyrki's comment above. The upshot for your problem is that a four dimensional representation is one of three things: the sum of four copies of the trivial irrep, or the sum of one copy of the trivial irrep with one of the two $3$ dimensional irreps.